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標題:

數學問題..急求幫忙?

發問:

懇請各位前輩較導..以下這兩條如何解答..看了教科書也不懂..本人數學真的很差.. Find all solutions of the equation for 0°≦t≦360° 1) 2cos^2 2t=-cos2t 2)sin^2 t - 4sint+1=0

最佳解答:

(1) 0° ≦ t ≦ 360° 0° ≦ 2t ≦ 720° 令 x = cos 2t , 則原式為: 2x^2 = - x 2x^2 + x = 0 x ( 2x + 1 ) = 0 當 x = cos 2t = 0 2t = 90° , 270° , 90°+360° , 270°+360° , 即: 2t = 90° , 270° , 450° , 630° t = 45° , 135° , 225° , 315° 當 x = cos 2t = - 1/2 2t = 120° , 240° , 120°+360° , 240°+360° , 即: 2t = 120° , 240° , 480° , 600° t = 60° , 120° , 240° , 300° Ans: t = 45° , 60° , 120° , 135° , 225° , 240° , 300° , 315° (2) 令 x = sin t , 則原式為: x^2 - 4x + 1 = 0 x = [ 4 ± √( 4^2 - 4 ) ] / 2 = ( 4 ± √12 ) / 2 =( 4 ± 2√3 ) / 2 = 2 ± √3 x = sin t 的值域為 [ - 1 , 1 ] , 故 x = 2 + √3 不合 所以 x = sin t = 2 - √3 t = arcsin ( 2 - √3 ) ≒ 15.54° , 180°-15.54° , 即: t ≒ 15.54° , 164.46° Ans: t ≒ 15.54° , 164.46°

其他解答:

1) 2cos^2 2t=-cos2t 設x=2t 2(cos x)^2=-cos x 2(cos x)^2 +cos x=0 (cos x)(2cos x+1)=0 cos x=0 or cos x=0.5 x=90 or x=270 or x=60 or x=-60 2t=90 or 2t=270 or 2t=60 or 2t=360-60 t=45 or t=135 or t=30 or t=150 2) sin^2 t - 4sint+1=0 設x=sin t x^2-4x+1=0 x=3.732050808 or x=0.267949192 sin t=3.732050808>1(delete) or sin t=0.267949192 t=15.54226822 or t=180-15.54226822=164.4577318C8D74AB62542840B